Thermal Expansion for fluid Questions in English

(B) As the temperature increases,the volume $V$ of a body increases due to thermal expansion,while its mass $m$ remains constant. Since density $\rho = m/V$,the density decreases.
Given the relation for volume expansion: $V = V_0(1 + \gamma \Delta \theta)$,where $\gamma$ is the coefficient of volume expansion.
The density at temperature $\theta$ is $\rho = m/V$ and at initial temperature $\theta_0$ is $\rho_0 = m/V_0$.
Therefore,$\frac{\rho}{\rho_0} = \frac{V_0}{V} = \frac{V_0}{V_0(1 + \gamma \Delta \theta)} = (1 + \gamma \Delta \theta)^{-1}$.
Using the binomial approximation for small $\gamma \Delta \theta$,we get $(1 + \gamma \Delta \theta)^{-1} \approx (1 - \gamma \Delta \theta)$.
Thus,$\rho = \rho_0(1 - \gamma \Delta \theta)$.

(A) The pressure exerted by a liquid column is given by $P = h \rho g$. Since the two columns balance each other,their pressures are equal: $h_1 \rho_1 g = h_2 \rho_2 g$,which implies $h_1 \rho_1 = h_2 \rho_2$.
Using the relation for density variation with temperature,$\rho = \frac{\rho_0}{1 + \gamma \theta}$,we have $\frac{h_1}{h_2} = \frac{\rho_2}{\rho_1} = \frac{1 + \gamma \theta_1}{1 + \gamma \theta_2}$.
Given $h_1 = 50 \ cm$,$\theta_1 = 50^{\circ}C$,$h_2 = 60 \ cm$,and $\theta_2 = 100^{\circ}C$.
Substituting these values: $\frac{50}{60} = \frac{1 + 50\gamma}{1 + 100\gamma}$.
Cross-multiplying: $50(1 + 100\gamma) = 60(1 + 50\gamma)$.
$50 + 5000\gamma = 60 + 3000\gamma$.
$2000\gamma = 10$.
$\gamma = \frac{10}{2000} = 0.005 \ ^{\circ}C^{-1}$.

(D) The relationship between volume and temperature at constant pressure is given by $V_t = V_0(1 + \gamma t)$,where $\gamma$ is the volume expansion coefficient.
For two different temperatures $t_1$ and $t_2$,we have $\frac{V_1}{V_2} = \frac{1 + \gamma t_1}{1 + \gamma t_2}$.
Given $V_1 = 100 \, cm^3$ at $t_1 = 20^{\circ}C$ and $V_2 = 125 \, cm^3$ at $t_2 = 100^{\circ}C$.
Substituting the values: $\frac{100}{125} = \frac{1 + 20\gamma}{1 + 100\gamma}$.
Simplifying the fraction: $0.8 = \frac{1 + 20\gamma}{1 + 100\gamma}$.
$0.8(1 + 100\gamma) = 1 + 20\gamma$.
$0.8 + 80\gamma = 1 + 20\gamma$.
$60\gamma = 0.2$.
$\gamma = \frac{0.2}{60} = \frac{1}{300} \approx 0.0033 \, ^{\circ}C^{-1}$.

(A) The relationship between density and temperature is given by $\rho = \frac{\rho_0}{1 + \gamma \Delta \theta} \approx \rho_0(1 - \gamma \Delta \theta)$.
Given:
$\rho_0 = 13.6 \, g/cm^3$ at $0^{\circ}C$ $(273 \, K)$.
$\gamma = 0.18 \times 10^{-3} \, ^{\circ}C^{-1}$.
Final temperature $T = 473 \, K$,which is $473 - 273 = 200^{\circ}C$.
So,$\Delta \theta = 200^{\circ}C - 0^{\circ}C = 200^{\circ}C$.
Substituting the values:
$\rho = 13.6 \times [1 - (0.18 \times 10^{-3} \times 200)]$
$\rho = 13.6 \times [1 - 0.036]$
$\rho = 13.6 \times 0.964 = 13.1104 \, g/cm^3$.
Thus,the density at $473 \, K$ is approximately $13.11 \, g/cm^3$.

(B) The apparent expansion of a liquid is the difference between its real expansion and the expansion of the container.
The volume expansion coefficient of the glass vessel $(\gamma_{vessel})$ is related to the linear expansion coefficient $(\alpha)$ by the formula $\gamma_{vessel} = 3\alpha$.
$\gamma_{vessel} = 3 \times 0.000009 = 0.000027 \text{ /}^{\circ}\text{C}$.
The apparent volume expansion coefficient $(\gamma_{app})$ is given by $\gamma_{app} = \gamma_{real} - \gamma_{vessel}$.
$\gamma_{app} = 0.000597 - 0.000027 = 0.00057 \text{ /}^{\circ}\text{C}$.

(C) Water exhibits anomalous expansion. It has maximum density at $4^{\circ}C$,which means it has minimum volume at this temperature.
If the water is heated above $4^{\circ}C$,its density decreases,causing its volume to increase.
If the water is cooled below $4^{\circ}C$,its density also decreases,causing its volume to increase.
Since the beaker is already completely filled at $4^{\circ}C$,any increase in volume due to heating or cooling will cause the water to overflow. Therefore,the water will overflow in both cases.

(A) The coefficient of apparent expansion $\gamma_{\text{app}}$ is defined by the formula:
$\gamma_{\text{app}} = \frac{\text{Mass expelled}}{\text{Mass remaining} \times \Delta T}$
Given that the mass expelled is $(1/100)$ of the mass remaining,let the mass remaining be $m$. Then the mass expelled is $m/100$.
The change in temperature $\Delta T = 80^{\circ}C$.
Substituting these values into the formula:
$\gamma_{\text{app}} = \frac{m/100}{m \times 80} = \frac{1}{100 \times 80} = \frac{1}{8000}$
$\gamma_{\text{app}} = 0.000125 = 1.25 \times 10^{-4} {^{\circ}C^{-1}}$
Thus,the correct option is $A$.

(C) The apparent expansion of mercury relative to the glass flask is given by the formula: $\Delta V = V_0 (\gamma_{real} - \gamma_{glass}) \Delta T$.
Since the coefficient of volume expansion of glass $\gamma_{glass} = 3\alpha_{glass}$,we have $\gamma_{glass} = 3 \times 9 \times 10^{-6} {^{\circ}C}^{-1} = 27 \times 10^{-6} {^{\circ}C}^{-1}$.
The change in temperature is $\Delta T = 38^{\circ}C - 18^{\circ}C = 20^{\circ}C$.
Substituting the values: $\Delta V = 50 \times (180 \times 10^{-6} - 27 \times 10^{-6}) \times 20$.
$\Delta V = 50 \times (153 \times 10^{-6}) \times 20 = 1000 \times 153 \times 10^{-6} = 0.153\, cc$.
Rounding to two decimal places,the amount of mercury above the mark is $0.15\, cc$.

(B) The volume of mercury that spills out is equal to the difference in the increase in volume of the mercury and the increase in the internal volume of the glass flask.
Given:
Initial volume $V_0 = 1000 \, cc$ (since $1 \, L = 1000 \, cc$).
Change in temperature $\Delta \theta = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.
Coefficient of volume expansion of mercury $\gamma_L = 1.82 \times 10^{-4} \, ^{\circ}C^{-1}$.
Coefficient of linear expansion of glass $\alpha_g = 0.1 \times 10^{-4} \, ^{\circ}C^{-1}$.
The coefficient of volume expansion of glass is $\gamma_g = 3\alpha_g = 3 \times 0.1 \times 10^{-4} = 0.3 \times 10^{-4} \, ^{\circ}C^{-1}$.
The volume of mercury that spills out is given by the formula:
$\Delta V = V_0 (\gamma_L - \gamma_g) \Delta \theta$
Substituting the values:
$\Delta V = 1000 \times (1.82 \times 10^{-4} - 0.3 \times 10^{-4}) \times 100$
$\Delta V = 1000 \times (1.52 \times 10^{-4}) \times 100$
$\Delta V = 15.2 \, cc$.
Thus,$15.2 \, cc$ of mercury will spill out.

(D) The change in volume of the mercury is given by $\Delta V = V_0 \gamma \Delta \theta$.
This change in volume causes the mercury to rise in the stem,so $\Delta V = A \Delta l$,where $A$ is the cross-sectional area and $\Delta l$ is the change in length.
Equating the two,we get $A \Delta l = V_0 \gamma \Delta \theta$.
Given: $V_0 = 10^{-6} \, m^3$,$\gamma = 18 \times 10^{-5} \, ^\circ C^{-1}$,$\Delta \theta = 100 \, ^\circ C$,and $A = 0.004 \, cm^2 = 0.004 \times 10^{-4} \, m^2 = 4 \times 10^{-7} \, m^2$.
Substituting the values: $\Delta l = \frac{V_0 \gamma \Delta \theta}{A} = \frac{10^{-6} \times 18 \times 10^{-5} \times 100}{4 \times 10^{-7}}$.
$\Delta l = \frac{18 \times 10^{-9}}{4 \times 10^{-7}} = 4.5 \times 10^{-2} \, m = 4.5 \, cm$.

(A) Let the initial height of the liquid in each arm at a reference temperature be $l_0$.
When the temperatures of the two arms are raised to $t_1$ and $t_2$ respectively,the new heights $l_1$ and $l_2$ are given by the thermal expansion formula $l = l_0(1 + \gamma \Delta t)$.
Assuming the reference temperature is $0^{\circ}C$,we have:
$l_1 = l_0(1 + \gamma t_1)$ and $l_2 = l_0(1 + \gamma t_2)$.
From these,we can write $l_0 = \frac{l_1}{1 + \gamma t_1} = \frac{l_2}{1 + \gamma t_2}$.
Equating the two expressions for $l_0$:
$l_1(1 + \gamma t_2) = l_2(1 + \gamma t_1)$
$l_1 + l_1 \gamma t_2 = l_2 + l_2 \gamma t_1$
$l_1 - l_2 = \gamma (l_2 t_1 - l_1 t_2)$
Therefore,the coefficient of volume expansion $\gamma$ is:
$\gamma = \frac{l_1 - l_2}{l_2 t_1 - l_1 t_2}$.

(D) The formula for volume expansion is $\Delta V = \gamma V_1 \Delta T$.
Here,$\Delta V = V_2 - V_1 = 125 \, cm^3 - 100 \, cm^3 = 25 \, cm^3$.
The initial volume $V_1 = 100 \, cm^3$.
The change in temperature $\Delta T = 100 \, ^\circ C - 20 \, ^\circ C = 80 \, ^\circ C$.
Substituting these values into the formula: $\gamma = \frac{\Delta V}{V_1 \Delta T} = \frac{25}{100 \times 80} = \frac{25}{8000} = 0.003125 \, ^\circ C^{-1}$.
Rounding to the given options,the value is $0.0031 \, ^\circ C^{-1}$.

(B) The volume of both the mercury and the glass flask increases with the rise in temperature.
The volume of mercury that overflows is the difference between the increase in the volume of mercury and the increase in the internal volume of the glass flask.
Given:
Initial volume $V_0 = 1 \, L = 1000 \, cm^3$
Change in temperature $\Delta \theta = 100 \, ^\circ C - 0 \, ^\circ C = 100 \, ^\circ C$
Coefficient of volume expansion of mercury $\gamma_l = 1.82 \times 10^{-4} \, ^\circ C^{-1}$
Coefficient of linear expansion of glass $\alpha_g = 0.1 \times 10^{-4} \, ^\circ C^{-1}$
Coefficient of volume expansion of glass $\gamma_g = 3 \alpha_g = 3 \times 0.1 \times 10^{-4} = 0.3 \times 10^{-4} \, ^\circ C^{-1}$
The volume of mercury that overflows is given by:
$\Delta V = V_0 (\gamma_l - \gamma_g) \Delta \theta$
$\Delta V = 1000 \times (1.82 \times 10^{-4} - 0.3 \times 10^{-4}) \times 100$
$\Delta V = 1000 \times (1.52 \times 10^{-4}) \times 100$
$\Delta V = 15.2 \, cm^3$

(C) Let $\rho_0$ and $\rho_T$ be the densities of glycerin at initial temperature and final temperature respectively.
The relationship between density and temperature is given by $\rho_T = \rho_0(1 - \gamma \Delta T)$,where $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in temperature.
Rearranging the equation,we get $\frac{\rho_T}{\rho_0} = 1 - \gamma \Delta T$.
The fractional change in density is defined as $\frac{\rho_0 - \rho_T}{\rho_0}$.
From the equation,$\frac{\rho_0 - \rho_T}{\rho_0} = \gamma \Delta T$.
Given $\gamma = 5 \times 10^{-4} \ K^{-1}$ and $\Delta T = 40 \ K$.
Substituting the values,the fractional change in density = $(5 \times 10^{-4} \ K^{-1}) \times (40 \ K) = 200 \times 10^{-4} = 0.02$.

(C) The density $\rho$ of a substance at temperature $T + \Delta T$ is related to its initial density $\rho_0$ at temperature $T$ by the formula: $\rho = \frac{\rho_0}{1 + \gamma \Delta T}$.
Since $\gamma \Delta T \ll 1$,we can use the binomial approximation: $\rho \approx \rho_0(1 - \gamma \Delta T)$.
The change in density is $\Delta \rho = \rho - \rho_0 = -\rho_0 \gamma \Delta T$.
The fractional change in density is given by $\frac{|\Delta \rho|}{\rho_0} = \gamma \Delta T$.
Given: $\gamma = 5 \times 10^{-4} \ K^{-1}$ and $\Delta T = 40^{\circ} C = 40 \ K$.
Therefore,the fractional change = $5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.02$.

(B) From the given graph,the volume of the gas at $t = 0 ^oC$ is $V_0 = 0.5 \text{ litre}$.
Using the formula for volume expansion: $V_t = V_0(1 + \alpha t)$.
Given $\alpha = \frac{1}{273} \text{ per } ^oC$ and $t = 819 ^oC$.
Substituting the values: $V_t = 0.5 \times (1 + \frac{1}{273} \times 819)$.
$V_t = 0.5 \times (1 + 3) = 0.5 \times 4 = 2 \text{ litre}$.
Since $1 \text{ litre} = 10^{-3} \text{ m}^3$,the volume is $2 \times 10^{-3} \text{ m}^3$.

(D) Let the initial volume of the vessel be $V_0$ and the coefficient of volume expansion for both the vessel and the oil be $\gamma$.
When the temperature increases by $\Delta T$,the new volume of the vessel is $V_v = V_0(1 + \gamma \Delta T)$.
The new volume of the oil is $V_o = V_0(1 + \gamma \Delta T)$.
Since the expansion is identical,the oil will continue to fill the vessel completely without any overflow.
Because the volume of the vessel has increased,it can now contain a larger volume of oil compared to the initial state.
However,since the density of the oil decreases as it expands,the mass $m = \rho V$ remains constant because the increase in volume is exactly compensated by the decrease in density.
Thus,the vessel contains a larger volume of oil,but the total mass of the oil remains the same.

(A) The coefficient of apparent expansion $\gamma_a$ is given by $\gamma_a = \gamma_l - \gamma_v$,where $\gamma_l$ is the real coefficient of volume expansion of the liquid and $\gamma_v$ is the coefficient of volume expansion of the vessel.
For the copper vessel: $C = \gamma_l - \gamma_c \implies \gamma_l = C + \gamma_c$.
For the silver vessel: $S = \gamma_l - \gamma_s$,where $\gamma_s$ is the coefficient of volume expansion of silver.
Substituting $\gamma_l$ from the first equation: $S = (C + \gamma_c) - \gamma_s$.
Rearranging for $\gamma_s$: $\gamma_s = C + \gamma_c - S$.
Since the coefficient of volume expansion is three times the coefficient of linear expansion $(\gamma_s = 3\alpha_s)$,we have $3\alpha_s = C + \gamma_c - S$.
Therefore,the coefficient of linear expansion of silver is $\alpha_s = \frac{C + \gamma_c - S}{3}$.

(C) Let $h_1 = 120 \, cm$ be the height of the liquid at $0^{\circ}C$ (left tube).
Let $h_2 = 120 + 4 = 124 \, cm$ be the height of the liquid at $30^{\circ}C$ (right tube).
Since the pressure at the bottom of both tubes must be equal:
$h_1 \rho_1 g = h_2 \rho_2 g$
$\rho_1 h_1 = \rho_2 h_2$
Since $\rho_2 = \frac{\rho_1}{1 + \gamma \Delta T}$,where $\Delta T = 30^{\circ}C - 0^{\circ}C = 30^{\circ}C$:
$h_1 = h_2 \left( \frac{1}{1 + \gamma \Delta T} \right)$
$1 + \gamma \Delta T = \frac{h_2}{h_1} = \frac{124}{120}$
$\gamma \Delta T = \frac{124}{120} - 1 = \frac{4}{120} = \frac{1}{30}$
$\gamma = \frac{1}{30 \times 30} = \frac{1}{900} \approx 0.00111 \, /^{\circ}C$
$\gamma \approx 11.11 \times 10^{-4} /^{\circ}C$.

(C) At the bottom of the $U$-tube,the pressure exerted by both liquid columns must be equal for equilibrium.
$P_T = P_0$
$\rho_T h_T g = \rho_0 h_0 g$
$\rho_T h_T = \rho_0 h_0 \quad \dots(1)$
The density of a liquid varies with temperature as $\rho_T = \frac{\rho_0}{1 + \gamma T}$,where $\gamma$ is the coefficient of volume expansion.
Substituting this into equation $(1)$:
$\left( \frac{\rho_0}{1 + \gamma T} \right) h_T = \rho_0 h_0$
$h_T = h_0 (1 + \gamma T)$
$h_T = h_0 + h_0 \gamma T$
$h_T - h_0 = h_0 \gamma T$
$\gamma = \frac{h_T - h_0}{h_0 T}$

(B) The coefficient of linear expansion of the container material is $\alpha$. Therefore,the coefficient of volume expansion of the container is $\gamma_{container} = 3 \alpha$.
When the system is heated,the liquid overflows if the increase in the volume of the liquid is greater than the increase in the volume of the container.
This implies that the coefficient of volume expansion of the liquid $(\gamma)$ must be greater than the coefficient of volume expansion of the container $(\gamma_{container})$.
Therefore,$\gamma > 3 \alpha$.

(C) The relationship between density $\rho$ and temperature $t$ is given by $\rho_t = \rho_0 (1 - \gamma \Delta t)$,where $\gamma$ is the coefficient of cubical expansion.
For a small change in temperature,the change in density is given by $\rho_2 = \rho_1 (1 - \gamma \Delta t)$.
Rearranging this,we get $\rho_2 - \rho_1 = -\rho_1 \gamma (t_2 - t_1)$.
Taking the magnitude,$\gamma = \frac{\rho_1 - \rho_2}{\rho_1 (t_2 - t_1)}$.
Given $\rho_1 = 0.998 \, g/cm^3$,$\rho_2 = 0.992 \, g/cm^3$,$t_1 = 20^oC$,and $t_2 = 40^oC$.
$\gamma = \frac{0.998 - 0.992}{0.998 \times (40 - 20)} = \frac{0.006}{0.998 \times 20} \approx \frac{0.006}{19.96} \approx 3.006 \times 10^{-4} \, ^oC^{-1}$.
Rounding to the nearest option,the value is $3 \times 10^{-4} \, ^oC^{-1}$.

(D) The real coefficient of expansion of a liquid is given by $\gamma_{\text{real}} = \gamma_{\text{app}} + \gamma_{\text{vessel}}$,where $\gamma_{\text{vessel}} = 3\alpha$ and $\alpha$ is the coefficient of linear expansion.
For vessel $A$,the real expansion is $\gamma_{\text{real}} = \gamma_1 + 3\alpha$.
For vessel $B$,the real expansion is $\gamma_{\text{real}} = \gamma_2 + 3\alpha_B$,where $\alpha_B$ is the coefficient of linear expansion of vessel $B$.
Since the liquid is the same,$\gamma_{\text{real}}$ is constant. Therefore,$\gamma_1 + 3\alpha = \gamma_2 + 3\alpha_B$.
Rearranging to solve for $\alpha_B$: $3\alpha_B = \gamma_1 - \gamma_2 + 3\alpha$.
Thus,$\alpha_B = \frac{\gamma_1 - \gamma_2}{3} + \alpha$.

(C) When a liquid is heated in a container,the change in volume of the liquid is $\Delta V_{\ell} = V \gamma \Delta T$ and the change in volume of the container is $\Delta V_{c} = V \gamma_{c} \Delta T$,where $\gamma_{c} = 3\alpha$.
For the liquid to overflow,the expansion of the liquid must be greater than the expansion of the container,i.e.,$\Delta V_{\ell} > \Delta V_{c}$.
Substituting the expressions,we get $V \gamma \Delta T > V (3\alpha) \Delta T$.
Dividing both sides by $V \Delta T$,we obtain $\gamma > 3\alpha$.

(B) The volume of the flask $V_0 = 1 \, L = 1000 \, cc$.
The change in temperature $\Delta \theta = 100 \, ^oC - 0 \, ^oC = 100 \, ^oC$.
The coefficient of volume expansion of mercury is $\gamma_m = 1.82 \times 10^{-4} / ^oC$.
The coefficient of linear expansion of glass is $\alpha_g = 0.1 \times 10^{-4} / ^oC$.
The coefficient of volume expansion of glass is $\gamma_g = 3 \alpha_g = 3 \times 0.1 \times 10^{-4} = 0.3 \times 10^{-4} / ^oC$.
The volume of mercury that spills out is the difference in the expansion of mercury and the expansion of the glass flask:
$\Delta V = V_0 (\gamma_m - \gamma_g) \Delta \theta$
$\Delta V = 1000 \times (1.82 \times 10^{-4} - 0.3 \times 10^{-4}) \times 100$
$\Delta V = 1000 \times (1.52 \times 10^{-4}) \times 100$
$\Delta V = 15.2 \, cc$.

(B) The coefficient of apparent expansion $\gamma_{app}$ is related to the coefficient of real expansion of liquid $\gamma_r$ and the coefficient of volume expansion of the vessel $\gamma_v$ by the formula: $\gamma_{app} = \gamma_r - \gamma_v$.
For a vessel,the coefficient of volume expansion $\gamma_v = 3\alpha$,where $\alpha$ is the coefficient of linear expansion.
Thus,$\gamma_{app} = \gamma_r - 3\alpha$.
For steel: $\gamma_{app, steel} = 114 \times 10^{-6} /^{\circ}C$ and $\alpha_{steel} = 12 \times 10^{-6} /^{\circ}C$.
$\gamma_r = \gamma_{app, steel} + 3\alpha_{steel} = 114 \times 10^{-6} + 3(12 \times 10^{-6}) = 114 \times 10^{-6} + 36 \times 10^{-6} = 150 \times 10^{-6} /^{\circ}C$.
For glass: $\gamma_{app, glass} = 132 \times 10^{-6} /^{\circ}C$.
Using $\gamma_{app, glass} = \gamma_r - 3\alpha_{glass}$:
$132 \times 10^{-6} = 150 \times 10^{-6} - 3\alpha_{glass}$.
$3\alpha_{glass} = 150 \times 10^{-6} - 132 \times 10^{-6} = 18 \times 10^{-6} /^{\circ}C$.
$\alpha_{glass} = 6 \times 10^{-6} /^{\circ}C$.

(D) The real coefficient of expansion of a liquid is constant and is given by the relation: $\gamma_{\text{real}} = \gamma_{\text{apparent}} + \gamma_{\text{vessel}}$.
For vessel $A$,the coefficient of volume expansion is $\gamma_{v,A} = 3\alpha$. Thus,$\gamma_{\text{real}} = \gamma_1 + 3\alpha$.
For vessel $B$,let the coefficient of linear expansion be $\alpha_B$. Then the coefficient of volume expansion is $\gamma_{v,B} = 3\alpha_B$. Thus,$\gamma_{\text{real}} = \gamma_2 + 3\alpha_B$.
Since $\gamma_{\text{real}}$ is the same for the liquid in both cases,we equate the two expressions:
$\gamma_1 + 3\alpha = \gamma_2 + 3\alpha_B$.
Rearranging for $\alpha_B$:
$3\alpha_B = \gamma_1 - \gamma_2 + 3\alpha$.
$\alpha_B = \frac{\gamma_1 - \gamma_2}{3} + \alpha$.

(C) The coefficient of cubical expansion of the container $(\gamma_c)$ is related to its coefficient of linear expansion $(\alpha)$ by the formula $\gamma_c = 3\alpha$.
Given that the coefficient of linear expansion of the container is $\alpha = \gamma/3$,we have $\gamma_c = 3 \times (\gamma/3) = \gamma$.
Since the coefficient of cubical expansion of the liquid $(\gamma_l = \gamma)$ is equal to the coefficient of cubical expansion of the container $(\gamma_c = \gamma)$,the volume of the liquid and the internal volume of the container increase by the same amount upon heating.
Therefore,the level of the liquid in the container will remain unchanged.

(C) The density $d$ of a substance at temperature $T$ is related to its initial density $d_i$ at temperature $T_0$ by the relation $d_f = \frac{d_i}{1 + \gamma \Delta T}$,where $\gamma$ is the coefficient of volume expansion and $\Delta T$ is the change in temperature.
The fractional change in density is given by $\frac{d_i - d_f}{d_i} = 1 - \frac{d_f}{d_i}$.
Substituting the expression for $d_f$,we get: $1 - \frac{1}{1 + \gamma \Delta T} = \frac{1 + \gamma \Delta T - 1}{1 + \gamma \Delta T} = \frac{\gamma \Delta T}{1 + \gamma \Delta T}$.
Since $\gamma \Delta T$ is very small $(5 \times 10^{-4} \times 40 = 0.02)$,we can approximate $1 + \gamma \Delta T \approx 1$.
Therefore,the fractional change $\approx \gamma \Delta T$.
Substituting the given values: $\text{Fractional change} = 5 \times 10^{-4} \times 40 = 200 \times 10^{-4} = 0.020$.

(C) The coefficient of volume expansion of glycerin is given as $\alpha_V = 49 \times 10^{-5} \; K^{-1}$.
The rise in temperature is $\Delta T = 30 \; ^{\circ}C$.
The relationship between volume and density is $\rho = \frac{m}{V}$,where $m$ is the mass and $V$ is the volume.
For a small change in temperature,the volume changes as $V' = V(1 + \alpha_V \Delta T)$.
The new density $\rho'$ is given by $\rho' = \frac{m}{V'} = \frac{m}{V(1 + \alpha_V \Delta T)} = \rho(1 + \alpha_V \Delta T)^{-1}$.
Using the binomial approximation $(1 + x)^{-1} \approx 1 - x$ for small $x$,we get $\rho' \approx \rho(1 - \alpha_V \Delta T)$.
The change in density is $\Delta \rho = \rho' - \rho = -\rho \alpha_V \Delta T$.
The fractional change in density is $\frac{|\Delta \rho|}{\rho} = \alpha_V \Delta T$.
Substituting the values: $\frac{|\Delta \rho|}{\rho} = 49 \times 10^{-5} \times 30 = 1470 \times 10^{-5} = 1.47 \times 10^{-2}$.

(N/A) Anomalous expansion of water:
Thermal expansion of water is non-uniform with temperature.
Water contracts on heating between $0^{\circ} C$ and $4^{\circ} C$. The volume of a given amount of water decreases as it is cooled from room temperature until its temperature reaches $4^{\circ} C$.
Below $4^{\circ} C$, the volume increases, and therefore the density decreases.
This means that water has a maximum density at $4^{\circ} C$.
This property has an important environmental effect: Bodies of water, such as lakes and ponds, freeze at the top first.
As a lake cools toward $4^{\circ} C$, water near the surface loses energy to the atmosphere, becomes denser, and sinks; the warmer, less dense water near the bottom rises. However, once the colder water on top reaches a temperature below $4^{\circ} C$, it becomes less dense and remains at the surface, where it freezes.
If water did not have this property, lakes and ponds would freeze from the bottom up, which would destroy much of their animal and plant life.

(N/A) The ideal gas equation is given by:
$PV = \mu RT$
Where $P$ is pressure,$V$ is volume,$\mu$ is the number of moles,$R$ is the gas constant,and $T$ is the absolute temperature.
At constant pressure,if the temperature changes by $\Delta T$,the volume changes by $\Delta V$:
$P(V + \Delta V) = \mu R(T + \Delta T)$
$PV + P\Delta V = \mu RT + \mu R\Delta T$
Since $PV = \mu RT$,we have:
$P\Delta V = \mu R\Delta T$
Dividing this equation by the original equation $PV = \mu RT$:
$\frac{P\Delta V}{PV} = \frac{\mu R\Delta T}{\mu RT}$
$\frac{\Delta V}{V} = \frac{\Delta T}{T}$
The coefficient of volume expansion $\alpha_V$ is defined as $\alpha_V = \frac{1}{V} \frac{\Delta V}{\Delta T}$.
Therefore,$\alpha_V = \frac{1}{T}$.
For an ideal gas,$\alpha_V$ depends on temperature and is inversely proportional to the absolute temperature $T$. As temperature increases,$\alpha_V$ decreases. At $0^{\circ}C$ $(273.15 \ K)$,$\alpha_V \approx 3.66 \times 10^{-3} \ K^{-1}$,which is significantly higher than that of solids and liquids.

(N/A) Volume expansion is the increase in the volume of a substance due to an increase in its temperature. When a solid,liquid,or gas is heated,its molecules vibrate more vigorously,causing the material to expand in all three dimensions.
The coefficient of volume expansion $(\gamma)$ is defined as the fractional change in volume per unit change in temperature.
Mathematically,it is expressed as: $\gamma = \frac{1}{V} \frac{\Delta V}{\Delta T}$,where $\Delta V$ is the change in volume,$V$ is the original volume,and $\Delta T$ is the change in temperature.
The $SI$ unit of the coefficient of volume expansion is per Kelvin $(K^{-1})$ or per degree Celsius $(^{\circ}C^{-1})$.

(N/A) $\alpha_V$ is defined as the coefficient of volume expansion. It represents the fractional change in volume per unit change in temperature.
Mathematically,$\alpha_V = \frac{1}{V} \frac{dV}{dT}$.
$\alpha_V$ depends on the nature of the material of the substance.
Its $SI$ unit is $\text{K}^{-1}$ or $^\circ\text{C}^{-1}$.

(A) The coefficient of volume expansion $\alpha_V$ represents the fractional change in volume per unit change in temperature.
Liquids generally expand more than solids,and among liquids,those with weaker intermolecular forces typically exhibit higher expansion coefficients.
Alcohol is an organic liquid with relatively weak intermolecular forces compared to the metallic bonding present in mercury.
Experimental values show that $\alpha_V$ for alcohol is approximately $1.1 \times 10^{-3} \ K^{-1}$,whereas for mercury,it is approximately $0.18 \times 10^{-3} \ K^{-1}$.
Therefore,the coefficient of volume expansion is greater for alcohol.

(N/A) The coefficient of volume expansion $\gamma$ is defined as $\gamma = \frac{1}{V} \left( \frac{dV}{dT} \right)$. At $4\,^{\circ}C$,water reaches its maximum density,which implies that its volume $V$ is at a minimum. Since the volume is at a minimum at this temperature,the rate of change of volume with respect to temperature $\left( \frac{dV}{dT} \right)$ is equal to $0$. Therefore,the coefficient of volume expansion for water at $4\,^{\circ}C$ is $0$.

(C) Water exhibits anomalous expansion. When water is heated from $0\,^{\circ}C$ to $4\,^{\circ}C$,its density increases,which means its volume decreases.
After reaching $4\,^{\circ}C$,the water behaves like a normal liquid,and its volume increases as it is heated from $4\,^{\circ}C$ to $10\,^{\circ}C$.
Therefore,the volume of water first decreases and then increases.

(N/A) The substance that contracts when its temperature increases is $Ice$ (between $0^{\circ}C$ and $4^{\circ}C$). This phenomenon is known as the anomalous expansion of water.

(A) For an ideal gas,the equation of state is $PV = nRT$.
At constant pressure,$V = (nR/P)T$.
The coefficient of volume expansion $\alpha_V$ is defined as $\alpha_V = \frac{1}{V} \left( \frac{dV}{dT} \right)_P$.
Differentiating $V$ with respect to $T$ at constant pressure,we get $\frac{dV}{dT} = \frac{nR}{P}$.
Substituting this into the definition,$\alpha_V = \frac{1}{V} \left( \frac{nR}{P} \right) = \frac{1}{V} \left( \frac{V}{T} \right) = \frac{1}{T}$.
At $0\,^{\circ}\text{C}$,the absolute temperature is $T = 273.15 \, \text{K}$.
Therefore,$\alpha_V = \frac{1}{273.15} \approx 3.66 \times 10^{-3} \, \text{K}^{-1}$.
Rounding to the standard approximation used in physics problems,we get $\alpha_V \approx 3.3 \times 10^{-3} \, \text{K}^{-1}$ (often cited as $1/273 \, \text{K}^{-1}$).

(A) For an ideal gas, the equation of state is $PV = nRT$.
At constant pressure, $P \Delta V = nR \Delta T$.
The coefficient of volume expansion $\alpha_{V}$ is defined as $\alpha_{V} = \frac{1}{V} \frac{dV}{dT}$.
From the ideal gas law, $\frac{dV}{dT} = \frac{nR}{P} = \frac{V}{T}$.
Therefore, $\alpha_{V} = \frac{1}{V} (\frac{V}{T}) = \frac{1}{T}$.
At room temperature $(T \approx 300 \ K)$, $\alpha_{V} = \frac{1}{300} \approx 3.33 \times 10^{-3} \ K^{-1}$.

(A) Let $V_{0}$ be the total capacity of the beaker and $V_{m}$ be the volume of mercury at $30^{\circ} C$. The unfilled volume is $\Delta V = V_{0} - V_{m}$.
When the temperature increases by $\Delta T$,the new capacity of the beaker is $V_{0}' = V_{0}(1 + \gamma_{b} \Delta T)$ and the new volume of mercury is $V_{m}' = V_{m}(1 + \gamma_{m} \Delta T)$.
The unfilled volume remains constant,so $\Delta V' = \Delta V$,which implies $V_{0}' - V_{m}' = V_{0} - V_{m}$.
Substituting the expressions for $V_{0}'$ and $V_{m}'$:
$V_{0}(1 + \gamma_{b} \Delta T) - V_{m}(1 + \gamma_{m} \Delta T) = V_{0} - V_{m}$
$V_{0} + V_{0} \gamma_{b} \Delta T - V_{m} - V_{m} \gamma_{m} \Delta T = V_{0} - V_{m}$
$V_{0} \gamma_{b} \Delta T = V_{m} \gamma_{m} \Delta T$
$V_{m} = \frac{V_{0} \gamma_{b}}{\gamma_{m}}$
Given $V_{0} = 500\, cc$,$\gamma_{b} = 6 \times 10^{-6}{ }^{\circ} C^{-1}$,and $\gamma_{m} = 1.5 \times 10^{-4}{ }^{\circ} C^{-1}$:
$V_{m} = \frac{500 \times 6 \times 10^{-6}}{1.5 \times 10^{-4}} = \frac{3000 \times 10^{-6}}{1.5 \times 10^{-4}} = \frac{3 \times 10^{-3}}{1.5 \times 10^{-4}} = 2 \times 10^{1} = 20\, cc$.

(B) The coefficient of real expansion of a liquid $(\gamma_r)$ is related to the apparent coefficient of expansion $(\gamma_a)$ and the coefficient of volume expansion of the vessel $(\gamma_v)$ by the relation: $\gamma_r = \gamma_a + \gamma_v$.
Let $\gamma_L$ be the coefficient of real volume expansion of the liquid.
For the brass vessel: $\gamma_L = X + 3\alpha$, where $3\alpha$ is the coefficient of volume expansion of brass.
For the tin vessel: $\gamma_L = Y + 3\alpha_{\text{tin}}$, where $\alpha_{\text{tin}}$ is the coefficient of linear expansion of tin.
Equating the two expressions for $\gamma_L$:
$X + 3\alpha = Y + 3\alpha_{\text{tin}}$
Rearranging to solve for $\alpha_{\text{tin}}$:
$3\alpha_{\text{tin}} = X + 3\alpha - Y$
$\alpha_{\text{tin}} = \frac{X + 3\alpha - Y}{3}$

(A) Given: Coefficient of volume expansion $\gamma = 49 \times 10^{-5} \, K^{-1}$ and change in temperature $\Delta T = 30^{\circ} C = 30 \, K$.
The density $\rho$ of a substance at temperature $T + \Delta T$ is given by $\rho_2 = \frac{\rho_1}{1 + \gamma \Delta T} \approx \rho_1(1 - \gamma \Delta T)$ for small values of $\gamma \Delta T$.
Rearranging the terms,we get $\frac{\rho_1 - \rho_2}{\rho_1} = \gamma \Delta T$.
The fractional change in density is $\frac{\Delta \rho}{\rho_1} = \gamma \Delta T$.
Substituting the values: $\frac{\Delta \rho}{\rho_1} = (49 \times 10^{-5}) \times 30 = 1470 \times 10^{-5} = 1.47 \times 10^{-2}$.

(C) Let $\alpha$ be the coefficient of linear expansion of the vessel. Given $\alpha = 1 \times 10^{-5} /^{\circ} C$.
The coefficient of cubical expansion of the vessel is $\gamma_{\text{vessel}} = 3\alpha = 3 \times 1 \times 10^{-5} /^{\circ} C = 3 \times 10^{-5} /^{\circ} C$.
Let $\gamma_{\text{Hg}}$ be the coefficient of cubical expansion of $Hg$.
For no overflow to occur when the vessel is heated,the expansion of the $Hg$ must be less than or equal to the expansion of the vessel's volume.
Therefore,$\gamma_{\text{Hg}} \leq \gamma_{\text{vessel}}$.
Substituting the value,we get $\gamma_{\text{Hg}} \leq 3 \times 10^{-5} /^{\circ} C$.

(C) The coefficient of cubical expansion of the vessel $(\gamma_v)$ is related to its coefficient of linear expansion $(\alpha)$ by the formula $\gamma_v = 3\alpha$.
Given that $\alpha = \frac{\gamma}{3}$,we have $\gamma_v = 3 \times (\frac{\gamma}{3}) = \gamma$.
Since the coefficient of cubical expansion of the liquid $(\gamma_l = \gamma)$ is equal to the coefficient of cubical expansion of the vessel $(\gamma_v = \gamma)$,both the liquid and the vessel expand by the same volume fraction for a given change in temperature.
Therefore,the level of the liquid in the vessel will remain almost the same.

(D) Water exhibits anomalous expansion between $0^{\circ} C$ and $4^{\circ} C$.
As water is heated from $0^{\circ} C$ to $4^{\circ} C$,its density increases,which means its volume decreases.
At $4^{\circ} C$,water attains its maximum density and minimum volume.
When water is heated further from $4^{\circ} C$ to $10^{\circ} C$,it expands like a normal liquid,and its volume increases.
Therefore,the volume of water first decreases and then increases.

(B) Volume expansion is the expansion of the volume of a gas due to an increase in its temperature.
$\therefore$ The coefficient of volume expansion $\alpha_{V}$ is defined as:
$\alpha_{V} = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_{p} ... (i)$
From the ideal gas equation,$pV = nRT$,we have $V = \frac{nRT}{p}$.
Differentiating with respect to $T$ at constant pressure $p$:
$\left( \frac{\partial V}{\partial T} \right)_{p} = \frac{nR}{p}$
Substituting this into equation $(i)$:
$\alpha_{V} = \frac{1}{V} \left( \frac{nR}{p} \right) = \frac{1}{V} \left( \frac{pV}{T} \cdot \frac{1}{p} \right) = \frac{1}{T}$
Thus,$\alpha_{V} = \frac{1}{T}$.
This implies that $\alpha_{V}$ is directly proportional to $\frac{1}{T}$. Therefore,a graph of $\alpha_{V}$ versus $\frac{1}{T}$ will be a straight line passing through the origin. This is correctly depicted in option $(b)$.

(B) Density is given by $\rho = \frac{m}{V}$.
Since mass $m$ remains constant,$\rho \propto \frac{1}{V}$.
Taking the logarithmic derivative,we get $\frac{d\rho}{\rho} = -\frac{dV}{V}$.
For small changes,$\frac{\Delta \rho}{\rho} = -\frac{\Delta V}{V}$.
We know that $\Delta V = V_0 \gamma \Delta T$,where $\gamma$ is the coefficient of volume expansion.
Thus,$\frac{\Delta \rho}{\rho} = -\gamma \Delta T$.
The percentage change in density is $\frac{\Delta \rho}{\rho} \times 100 = -\gamma \Delta T \times 100$.
Given $\gamma = 18.2 \times 10^{-5} \ K^{-1}$ and $\Delta T = 60^{\circ} C - 10^{\circ} C = 50 \ K$.
Percentage change $= 18.2 \times 10^{-5} \times 50 \times 100 = 18.2 \times 10^{-5} \times 5000 = 0.91 \%$.

(A) The volume of the liquid that overflows is equal to the difference between the expansion of the liquid and the expansion of the glass vessel.
Increase in volume of liquid,$\Delta V_L = V_0 \gamma_L \Delta T$.
Increase in volume of the glass vessel,$\Delta V_g = V_0 \gamma_g \Delta T$.
Since the coefficient of volume expansion $\gamma_g = 3\alpha_g$,we have $\Delta V_g = V_0 (3\alpha_g) \Delta T$.
The volume of liquid that overflows is $\Delta V_{overflow} = \Delta V_L - \Delta V_g$.
$\Delta V_{overflow} = V_0 \gamma_L \Delta T - V_0 (3\alpha_g) \Delta T$.
$\Delta V_{overflow} = V_0 \Delta T (\gamma_L - 3\alpha_g)$.

(B) The volume of mercury that overflows is equal to the difference in the volume expansion of mercury and the glass flask.

$\Delta V = V_0 (\gamma_m - \gamma_g) \Delta \theta$

Given that $\gamma_g = 3 \alpha_g$, where $\alpha_g$ is the coefficient of linear expansion of glass.

$\gamma_g = 3 \times (0.1 \times 10^{-4} /{ }^{\circ} C) = 0.3 \times 10^{-4} /{ }^{\circ} C = 30 \times 10^{-6} /{ }^{\circ} C$

Given $\gamma_m = 1.82 \times 10^{-4} /{ }^{\circ} C = 182 \times 10^{-6} /{ }^{\circ} C$, $V_0 = 1 \text{ litre} = 1000 \text{ cc}$, and $\Delta \theta = 100^{\circ} C$.

$\Delta V = 1000 \times (182 \times 10^{-6} - 30 \times 10^{-6}) \times 100$

$\Delta V = 1000 \times (152 \times 10^{-6}) \times 100 = 15.2 \text{ cc}$

Thus, the amount of mercury that overflows is $15.2 \text{ cc}$.